Root of f(x) = x^3-x-1 ?

How does one find the root of f(x) = x^3 - x - 1 ? Quadratic Equation only works on power of 2. I can't factor out an x to get a first term of x^2 because then Quadratic equation still won't work because the middle and last term would be messed up, I think.

Staff: Mentor

There is a formula, like the one you've memorized for quadratics, though the one for cubics is more frightening. Try a google search for it.

Or you can sometimes see a solution by inspection, esp. integer solutions. Or you can use a numerical method to find an approximate solution, that approximation can be to whatever accuracy you desire. Or you could plot a graph and read off from there.

The rational root test won't give a root here because this polynomial has NO rational roots. The rational root test, for this polynomial, says that 1 and -1 are the only possible rational roots and they clearly are not roots.

If a and b are any two numbers then [itex](a- b)^3= a^3- 3a^2b+ 3ab^2- b^3[/itex] and [itex]3ab(a- b)= 3a^2b- 3ab^2[/itex] so that [itex](a- b)^3+ 3ab(a- b)= a^3- b^3[/itex]. That means that if we let x= a- b, m= 3ab, and [itex]n= a^3- b^3[/itex], we have [itex]x^3+ mx= n[/itex].

Now, suppose we know m and n. Can we solve for a and b and so find x? Yes, we can!

From m= 3ab, we have b= m/3a. Putting that into [itex]n= a^3- b^3[/itex], we get [itex]n= a^3- m^3/3^3a^3[/itex]. Multiplying through by [itex]a^3[/itex] we have [itex]na^3= (a^3)^2- m^3/3^3[/itex] which is a quadratic [itex](a^3)^2- na^3- m^3/3^3= 0[/itex] for [itex]a^3[/itex].

Solving that with the the quadratic formula,
[tex]a^3= \frac{n\pm\sqrt{n^2+4\frac{m^3}{3^3}}}{2}= \frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}[/tex]

The rational root test won't give a root here because this polynomial has NO rational roots. The rational root test, for this polynomial, says that 1 and -1 are the only possible rational roots and they clearly are not roots.

You are right that in this case it wouldn't help, but it is a good rule of thumb in general.